初中数学八年级下册（648题）

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探究：\(CE\)与\(CG\)有怎样的位置关系？请说明理由。

 

知识点：复习

参考答案：\(CE \bot CG\)，
理由如下：
\(\because \)正方形\(DEFG\)和正方形\(ABCD\)，
\(\therefore DE = DG\)，\(AD = DC\)，
\(\because \angle CDG + \angle CDE \)
\(= \angle ADE + \angle CDE = 90^\circ \)，
\(\therefore \angle CDG = \angle ADE\)，
在\(\Delta ADE\)和\(\Delta CDG\)中，
\(\left\{ {\begin{array}{*{20}{l}}
{AD = CD} \\
{\angle ADE = \angle CDG} \\
{DE = DG}
\end{array}} \right.\)，
\(\therefore \Delta ADE \cong \Delta CDG(SAS)\)，
\(\therefore \angle CDA = \angle DCG\)，
\(\because \angle ACD + \angle CAD + \angle ADC = 180^\circ \)，
\(\angle ADC = 90^\circ \)，
\(\therefore \angle ACG = \angle ACD + \angle DCG \)
\(= \angle ACD + \angle CAD = 90^\circ \)，
\(\therefore CE \bot CG\)。